Chapter 9.6, Problem 4P

a)

To determine

To evaluate: the solutions of $6x^2-5x=7$ .

Answer to Problem 4P

The solution of above system of equations is (3, 8) and (−1, 4).

Explanation of Solution

Given:

  $6x^2-5x=7$

Using the quadratic formula,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

From the equation,

  $6x^2-5x=7$

finding the value of a , b and c ,

  $\begin{array}{l}a=6\\ b=-5\\ c=-7\end{array}$

Using quadratic formula to solve above quadratic equation,

  $\begin{array}{c}x=\frac{5\pm \sqrt{{\left(-5\right)}^2-4\left(6\right)\left(-7\right)}}{2\left(1\right)}\\ =\frac{5\pm \sqrt{25+168}}{2}\\ =\frac{5\pm \sqrt{193}}{2}\end{array}$

So, the given equation has two solutions.

b)

To determine

To evaluate: the number of real solutions of the equation $a{x}^2+bx+c=0$ if a is positive and c is negative and explain.

Answer to Problem 4P

There will be two real number solutions.

Explanation of Solution

Given:

  $a{x}^2+bx+c=0$

Using the discriminant formula,

  $d=\sqrt{b^2-4ac}$

From the equation, $a{x}^2+bx-c=0$ finding the value of a , b and c ,

a = a ,

b = b ,

c = − c ,

Using quadratic formula to solve above quadratic equation,

  $\begin{array}{c}d=\sqrt{b^2-4\left(a\right)\left(-c\right)}\\ =\sqrt{b^2+4ac}\end{array}$

It can be seen that the value of discriminant is greater than or equal to zero.

It is known that when the value of discriminant is greater than or equal to zero, then quadratic equation will have two real-number solutions.